60% of a first order reaction was completed in 60 minutes. When was 50% of reaction completed?

45.22 minutes

For first order reaction,

k = \(\frac{2.303}{t}\)log\(\frac{a}{a-x}\)

k = \(\frac{2.303}{60}\)log\(\frac{100}{40}\)

k = \(\frac{2.303}{60}\)log\(\frac{5}{2}\)

k = \(\frac{2.303}{60}\)(log5 - log2)

k = \(\frac{2.303}{60}\)(0.69 - 0.30)

k = \(\frac{2.303}{60}\) x 0.39

k = 0.0149695

t_1/2 = t_50% = \(\frac{0.693}{k}\) = \(\frac{0.693}{0.0149695}\) ≈ 45.22 minutes